Nonregular languages

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Non-regular languages

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Non-regular languages
Regular languages
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How can we prove that a language is not regular?
Problem this is not easy to prove
Solution the Pumping Lemma !!!
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The Pigeonhole Principle

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pigeons

pigeonholes
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A pigeonhole must contain at least two pigeons
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pigeons
...........
pigeonholes
...........
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The Pigeonhole Principle

pigeons
pigeonholes
There is a pigeonhole with at least 2 pigeons
...........
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The Pigeonhole Principleand DFAs

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DFA with states
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In walks of strings
no state is repeated

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In walks of strings
a state is repeated

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If string has length

Then the transitions of string are more than the
states of the DFA
Thus, a state must be repeated
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In general, for any DFA String has
length number of states
A state must be repeated in the walk of
walk of
......
......
Repeated state
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In other words for a string
transitions are pigeons
states are pigeonholes
walk of
......
......
Repeated state
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The Pumping Lemma

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Take an infinite regular language
There exists a DFA that accepts
states
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Take string with
There is a walk with label
.........
walk
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If string has length
(number of states of DFA)
then, from the pigeonhole principle
a state is repeated in the walk
......
......
walk
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Let be the first state repeated in the walk
of
......
......
walk
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Write
......
......
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Observations
length
number of states of DFA
length
......
......
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The string is accepted
Observation
......
......
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The string is accepted
Observation
......
......
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The string is accepted
Observation
......
......
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The string is accepted
In General
......
......
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In General
Language accepted by the DFA
......
......
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In other words, we described
The Pumping Lemma !!!
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The Pumping Lemma

• Given a infinite regular language

• there exists an integer

• for any string with length

• we can write

• with and

• such that

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Applications ofthe Pumping Lemma

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Theorem
The language
is not regular
Proof
Use the Pumping Lemma
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Assume for contradiction that is a regular
language
Since is infinite we can apply the Pumping
Lemma
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Let be the integer in the Pumping Lemma
Pick a string such that
length
We pick
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Write
From the Pumping Lemma
it must be that length
Thus
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From the Pumping Lemma
Thus
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From the Pumping Lemma
Thus
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BUT
CONTRADICTION!!!
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Our assumption that is a regular language is not
true
Therefore
Conclusion
is not a regular language
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Non-regular languages
Regular languages
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Non-regular languages
Regular languages
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Theorem
The language
is not regular
Proof
Use the Pumping Lemma
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Assume for contradiction that is a regular
language
Since is infinite we can apply the Pumping
Lemma
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Let be the integer in the Pumping Lemma
Pick a string such that
and
length
We pick
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Write
From the Pumping Lemma
it must be that length
Thus
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From the Pumping Lemma
Thus
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From the Pumping Lemma
Thus
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BUT
CONTRADICTION!!!
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Our assumption that is a regular language is not
true
Therefore
Conclusion
is not a regular language
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Non-regular languages
Regular languages
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Theorem
The language
is not regular
Proof
Use the Pumping Lemma
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Assume for contradiction that is a regular
language
Since is infinite we can apply the Pumping
Lemma
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Let be the integer in the Pumping Lemma
Pick a string such that
and
length
We pick
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Write
From the Pumping Lemma
it must be that length
Thus
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From the Pumping Lemma
Thus
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From the Pumping Lemma
Thus
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BUT
CONTRADICTION!!!
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Our assumption that is a regular language is not
true
Therefore
Conclusion
is not a regular language
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Non-regular languages
Regular languages
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Theorem
The language
is not regular
Proof
Use the Pumping Lemma
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Assume for contradiction that is a regular
language
Since is infinite we can apply the Pumping
Lemma
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Let be the integer in the Pumping Lemma
Pick a string such that
length
We pick
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Write
From the Pumping Lemma
it must be that length
Thus
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From the Pumping Lemma
Thus
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From the Pumping Lemma
Thus
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Since
There must exist such that
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However
for
for any
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BUT
CONTRADICTION!!!
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Our assumption that is a regular language is not
true
Therefore
Conclusion
is not a regular language
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Summary

• Showing regular
• construct DFA, NFA
• construct regular expression
• show L is the union, concatenation, intersection
  (regular operations) of regular languages.
• Showing non-regular
• pumping lemma
• assume regular, apply closure properties of
  regular languages and obtain a known non-regular
  language.